I need to prove that the sequence $\{x_n\}_{n=1}^\infty $ converges, where $x_1=1$ and $$x_{n+1}=(\sqrt{2})^{x_n} $$ The only way I can see to progress is to use the monotone bounded sequence theorem. (Nothing else makes sense to me.) My intuition tells me the sequence should be bounded above by $2$, and that $x_{n+1}>x_n$, but I can't work out what to do to prove those.
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1$\begingroup$ At the end, you want $x_{n+1}\gt x_n$. We haven't heard about $a$ before and you have the sense of the inequality wrong. You are correct that it is bounded above by $2$. Assume $x_n \lt 2$ and use the monotonicity of $\sqrt 2^x$ to conclude $x_{n+1} \lt 2$ $\endgroup$– Ross MillikanApr 17, 2018 at 22:42
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$\begingroup$ sorry, dont know why i changed to $a_n$, I'll fix that. $\endgroup$– BobbieApr 17, 2018 at 23:07
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1 Answer
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If $x_n < 2$, then
$$x_{n + 1} = \sqrt{2}^{x_n} < \sqrt{2}^2 = 2.$$
Furthermore, the function $x \mapsto 2^{x/2}$ is a strictly increasing function (its derivative is $2^{x/2} \cdot \ln \sqrt 2 > 0$, for example), and so $x_{n + 1} > x_n$.